A plane is flying with an airspeed of 170 miles per hour and heading 150°. The wind currents are running at 30 miles per hour at 170° clockwise from due north. Use vectors to find the true course and ground speed of the plane. (Round your answers to the nearest ten for the speed and to the nearest whole number for the angle.)
I get a ground speed of $199.6$ mi/h at a bearing of $151°$ from due north.
I think my bearing is off a bit.
Draw a diagram and denote $O$ = the origin, and $A$ be the point in the IV quadrant such that $OA = 170$ and $<(Ox,OA) = 60$ degrees, and $B$ be another point also in the IV quadrant such that $AB = 30$ , and $<(AO,AB) = <OAB = 160$ degrees. So vectors $OA$, and $AB$ represent the plane's airspeed and the wind's speed respectively. Thus the vector $OB$ is the plane's velocity relative to the ground, and its magnitude is the plane's ground speed. By the law of cosine:
$OB = \sqrt{OA^2 + AB^2 - 2\cdot OA\cdot AB\cdot cos160} = \sqrt{170^2 + 30^2 - 2\cdot 170\cdot 30\cdot cos160} = 198.45$ mph.
And by the law of sine in $\triangle OAB$: $\dfrac{AB}{sin<AOB} = \dfrac{OB}{sin<OAB}$:
Then: $<AOB = sin^{-1}\left(\dfrac{AB\cdot sin<OAB}{OB}\right) = sin^{-1}\left(\dfrac{30\cdot sin160}{198.45}\right) = 2.96$ degrees.
So: $<(Ox,OB) = 60 + 2.96 = 62.96$ degrees. Thus the true course of the plane is: $90 + 62.96 = 152.96$ degrees due north.